The 2D MHD Systems with Vertical Dissipation and Vertical Dissipation Magnetic Diffusion

(1)

http://www.scirp.org/journal/ajcm

ISSN Online: 2161-1211 ISSN Print: 2161-1203

DOI: 10.4236/ajcm.2019.92007 Jun. 26, 2019 81 American Journal of Computational Mathematics

The 2D MHD Systems with Vertical Dissipation

and Vertical Dissipation Magnetic Diffusion

Xiaoting Yang

Department of Mathematics, Jinan University, Guangzhou, China

Abstract

In this paper, we study the global regularity of the classical solution of the 2D incompressible magnetohydrodynamic equation with vertical dissipation and vertical magnetic dissipation. We show that any solution of the second component

(

u b2, 2

)

has a global

2r

L -bound, where r satisfies 1≤ < ∞r

and the boundary does not grow faster than rlogr as r increases.

Keywords

MHD Equation, Global Regularity, Vertical Dissipation

1. Introduction

The generalized MHD system is

2

, , 0.

t t

u u u p b b u

b u b b u b

u b

α β

ν κ

+ ⋅∇ = −∇ + ⋅∇ − Λ + ⋅∇ = ⋅∇ − Λ ∇ ⋅ = ∇ ⋅ =

(1)

where ν κ α β, , , >0,

( )

1 2

Λ = −∆ _{, u denotes the velocity field and b denotes the} magnetic field. The magnetohydrodynamic (MHD) systems [1] control the dynamics of velocity and magnetic fields in conductive fluids such as plasma and reflect the basic laws of physical conservation.

In recent years, the MHD equations with partial dissipation regularity problem have attracted considerable interests. For example, the n-dimensional MHD Equation (1), when the coefficient satisfies

1

, 0, 1 ,

2 4 2

n n

α ≥ + β> α β+ ≥ +

it has been proved that the solution has global regularity [2]. Wu [3] has been

How to cite this paper: Yang, X.T. (2019) The 2D MHD Systems with Vertical Dissi-pation and Vertical DissiDissi-pation Magnetic Diffusion. American Journal of Computa-tional Mathematics, 9, 81-96.

https://doi.org/10.4236/ajcm.2019.92007

Received: May 9, 2019 Accepted: June 23, 2019 Published: June 26, 2019

http://creativecommons.org/licenses/by/4.0/

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DOI: 10.4236/ajcm.2019.92007 82 American Journal of Computational Mathematics proved the 2D GMHD admits a global regularity for a three-case:

1 1

, 1; 0 , 2 2; 2, 0.

2 2

α≥ β ≥ ≤ <α α β+ > α ≥ β=

And it is also proved that the condition satisfying ν=0, 1β> has a global smooth solution with the direction of the magnetic field that remains sufficiently smooth. Cao, Regmi and Wu [4] have been proved that the 2D MHD with horizontal dissipation and horizontal magnetic diffusion in horizontal component of any solutions has a global regularity. The global regularity of the class solution of the MHD equation with magnetic diffusion and mixed partial dissipation is established by Wu [5]. In [6], the global existence and uniqueness of the smooth solution of 2D micropolar fluid flow with zero angular viscosity have been proved. Other related articles can be seen in [7][8][9], etc.

In this paper, we study the 2D MHD systems with vertical dissipation and vertical dissipation magnetic diffusion, namely

2 2 2 2

, ,

0, 0.

t t

u u u p u b b

b u b b b u

u b

+ ⋅∇ = −∇ + ∂ + ⋅∇ + ⋅∇ = ∂ + ⋅∇

∇ ⋅ = ∇ ⋅ =

(2)

In this case, we only get the global 2r

L -bound of the solution in the y-direction,

and the global regularity problem for the complete directional solution has not been achieved.

In the following article, let w±= ±u b, this will provide us with convenience.

We have a symmetric equation by (2), namely

(

)

(

)

2 2 2 2

,

0.

t

w w w p w

w w

+ − + +

− + − −

+ −

∂ + ⋅∇ = −∇ + ∂

∇ ⋅ = ∇ ⋅ =

(3)

The new Equation (3) consists of two vectors, which is more complicated in the calculation process, therefore, we use fractionally derivative triple product estimation [4] to solve this difficulty. This paper takes Cao and Wu recent study of two-dimensional partially dissipated Boussinesq equation [8] as an example to discuss the influence of known vertical component

(

u b2, 2

)

Lebesgue norm on

global regularity. And in Section 4, we obtain the main Theorem 3, which proves that

(

u b2, 2

)

₂_r ≤C rlogr for 2< < ∞r . In fact, in Section 2 we get

Theorem 1, which is about the solution of Equation (2) bounded by Lebesgue in the y-direction. The sameness of Theorem 1 and Theorem 3 is that boundedness is related to the r, but in Theorem 1, we get the case of r=1, and Theorem 3 has a slower bounded change with the increase of r.

The rest of this article is divided into four parts. In Section 2, we prove the global bounded for

(

u b2, 2

)

₂_r, and the boundedness depends on the index of r. In Section 3, we show the global bounded for pq and

( )

2

0 sd

T

H

p τ τ

∫

with

( )

0,1

s∈ . In Section 4, we prove that the solution of (2) in y-direction has a

global Lebesgue bound. In Section 5, we prove the bounded condition of

(

u b2, 2

)

under the

2

t y

(3)

DOI: 10.4236/ajcm.2019.92007 83 American Journal of Computational Mathematics

2. A Global Bound in the Lebesgue Spaces

In this section, we prove the classical solution of (2) at the y-direction exists globally bounded in 2r

L norm. The boundedness obtained here depends on the

index of r. We have the following theorem.

Theorem 1. Assume that

(

)

2

( )

2

0, 0

u b ∈H  and ut=0=u b0, t=0=b0 ,

( )

u b, be the corresponding solution of (2). For any 1≤ < ∞r ,

(

u b2, 2

)

obeys

global bound

(

)

3

2

2, 2 ₂ 1e ,

C r r

u b ≤C (4)

where C1 and C2 are constants depending on

(

u b0, 0

)

₂_r only.

To prove the Theorem 1, we need to estimate the global bounded under 2 L

norm.

Lemma 1. Let

(

)

2

( )

2

0, 0

u b ∈H  and let

( )

u b, be the corresponding solution of (2). Then, for ant t≥0,

( )

u b, obeys the

( )

2

( )

2

( )

2

( )

2 2 2

2 2 0 2 0 2

2 2 2 0 2d 2 0 2d .

t t

u t + b t +

∫

∂ u τ τ+

∫

∂ b τ τ≤ u + b

Here we omit the proof of Lemma 1 and now begin to prove Theorem 1. Proof. Taking the product of the second component of the first equation of (3)

with 2 2

2 2

r

w w+ + − , and integrating with respect to space variable, we obtain

(

)

(

)

2 2 2 2 2 2

2 ₂ 2 2 2 2 2 2

1 d

2 1 d 2 1 d ,

2 d

r r r

r

w r w w x r p w w x

r t

− −

+ ₊ ₋ _∂ + + ₌ ₋ _∂ + +

∫

(5)

note that

(

)

2 2

2 2 2 d 0.

r

w−⋅∇ w w w+ + + − x=

∫

By Hölder’s and Sobolev’s inequalities, and using Young’s inequality, we got

(

)

( )

2 2

2 2 2

1 1

2

2 2 2 2

2

2 1

1 1

2 ₂ ₂ ₂ ₂

2 2 1

2 1

2

1 3 2 ₂

2

2 1 2 2

2 1

2 1 d

,

2 1

, 2

r

r r

r

r r

r

r r

r

r p w w x

p w w w

Cr p w w w

r

w w Cr p w

− + +

− −

+ + +

−

− −

+ + +

+

− −

+ + +

+

− ∂

≤ ∇ ∂

−

≤ ∂ + ∇

∫

where C is a constant independent of r. In order to bound the pressure, we take the divergence of (3), we get

(

)

(

)

1 2 2 1 2 2 1 2 2 2 2 2 2 2 .

p w− w+ w+ w− w− w+ w+ w−

−∆ = ∂ ∂ + ∂ + ∂ ∂ + ∂ (6)

Since, the Riesz transform [10] has bounds for any 1< < ∞p on Lp, we have

(

)

(

)

2 2 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2

1 1 1 1 1

2 ₂ 2 1 ₂ 2 2 ₂ 2 ₂ 2 1 ₂ 2 2 ₂ .

r r r r r

r r

p w w w w w w w w

w w w w w w

− + + − − + + −

+ + + + +

− + + + − −

∇ ≤ ∂ + ∂ + ∂ + ∂

≤ ∂ + ∂ + ∂ + ∂ (7)

(4)

( )

(

)

(

)

( )

(

)(

)

2 1 2

2 ₂

2 1

2

2 2 2 2 2 1

2 1 ₂ 2 2 ₂ 2 1 ₂ 2 2 ₂ 2 ₂ 2 ₂ 2 ₂

2 2 2 2 2 2

2 ₂ 2 ₂ 2 1 ₂ 2 2 ₂ 2 1 ₂ 2 2 ₂ .

r r

r

r r r

r r

p w

w w w w w w w

w w w w w w

− + +

−

+ + − − − + +

− + + + − −

∇

≤ ∂ + ∂ + ∂ + ∂ +

≤ + ∂ + ∂ + ∂ + ∂

Based on the above estimates, we get

(

)

(

)(

)

2

2 1

2 ₂ 2 2 2

2

2 2 2 2 2 2

3

2 ₂ 2 ₂ 2 1 ₂ 2 2 ₂ 2 1 ₂ 2 2 ₂

1 d

2 1

d

.

r r

r

r r

w r w w

r t

Cr w w w w w w

−

+ + +

− + + + − −

+ − ∂

≤ + ∂ + ∂ + ∂ + ∂

Similarly,

(

)

(

)(

)

2

2 1

2 ₂ 2 2 2

2

2 2 2 2 2 2

3

2 ₂ 2 ₂ 2 1 ₂ 2 2 ₂ 2 1 ₂ 2 2 ₂

1 d

2 1

d

.

r r

r

r r

w r w w

r t

Cr w w w w w w

−

− − −

− + + + − −

+ − ∂

≤ + ∂ + ∂ + ∂ + ∂

Combine these two inequalities to get

(

)

(

)

(

)(

)

2 2

2 2 1 1

2 ₂ 2 ₂ 2 2 2 2 2 2

2 2

2 2 2 2 2 2

3

2 2 2 1 2 2 2 1 2 2

2 2 2 2 2 2

1 d

2 1

d

.

r r r r

r r

w w r w w w w

r t

Cr w w w w w w

− −

+ − + + − −

− + + + − −

 

+ + − _ ∂ + ∂ _

 

≤ + ∂ + ∂ + ∂ + ∂

Following the Gronwall's inequality, we obtain

( )

(

)

(

)

(

)

2 2

2 ₂ 2 ₂

2 2

2 ₂ 2 ₂

2 2 2 2

4

2 1 2 2 2 1 2 2

0 2 2 2 2

0 0

exp d .

r r

t

w w

Cr w w w w τ

+ −

+ + − −

+

≤ +

×

_∫

∂ + ∂ + ∂ + ∂

According to Lemma 1, get (4). □

3. Global Bounds for the Pressure

In this section, we show the solution of the first components

(

u b1, 1

)

has a 2 L

-bound with r=2 or r=3, and establish the pressure has a global bound.

The results can be stated as follows.

(

)

2

( )

2

0, 0

( )

u b, be the corresponding solution of (2)

(

u b1, 1

)( )

t ₂_r≤C, r=2, 3, (8)

for any T >0, and t≤T,

( )

2

0

, sd ,

T

q H

p t ≤C

∫

p τ τ≤C (9)

where 1< ≤q 3 and s∈

( )

0,1 , and C is a constant related to T and initial value. Here we use two calculus inequalities of the following lemma.

Lemma 2. [4] Assume that 2

( )

2

(5)

DOI: 10.4236/ajcm.2019.92007 85 American Journal of Computational Mathematics then

1 1

2 2

1 2

4 1 2 ,

f ≤C ∂ f ∂ f (10)

1 1 1

3 3 3

1 2

3 2 1 2.

f ≤C f ∂ f ∂ f (11)

Proof. We use the symmetric Equation (3) to prove the case of r=2 in Theorem 2. Take the inner product of the first Equation (3) with 2

1 1

w w+ + , we

obtain

4 2 2 2

1 ₄ 2 1 1 1 1 1

1 d

3 d 3 d .

4 dt w w w x p w w x

+ ₊ _∂ + + ₌ _∂ + +

∫

(12)

Using ∇ ⋅w+=0 and integrate by parts, we get 2

1 1 1 2 2 2 1

2

2 2 1 2 2 1 1

1 2

d

d 2 d

2 ,

p w w x

p w w x

pw w x pw w w x

I I

+ + + +

+ + + + +

∂

= − ∂

= ∂ + ∂

= +

∫

by Hölder’s and Sobolev’s inequalities,

2 2 2 1 1 4 2 4 2 1 1 2

4 ₂ ₁ ₂ ₁

4 2

3

d

.

I pw w w x p w w w

C p w w w

+ + + + + +

+ + +

= ∂ ≤ ∂

≤ ∇ ∂

∫

(13)

According to (7),

(

)

(

)

4 ₂ ₂ ₁ ₂ ₂ ₂ ₂ ₁ ₂ ₂

4 2 2 4 2 2

3

.

p w− w+ w+ w+ w− w−

∇ ≤ ∂ + ∂ + ∂ + ∂

Therefore, by Young’s inequality,

(

)

(

)

2 4 4

2 1 2 1 ₂ 2 ₄ 2 ₄

2 2 2 2

2 1 ₂ 2 2 ₂ 2 1 ₂ 2 2 ₂ 1

3

.

I w w C w w

w w w w

+ + − +

+ + − −

≤ ∂ + +

× ∂ + ∂ + ∂ + ∂

To bound I1, we first apply Hölder inequality,

( )

2 8

1 2 2 ₈ 1

4 5

.

I ≤ ∂ p w+ w+ (14)

According to Lemma 2 and ∇ ⋅w+ =0,

( )

1

( )

1 1 1

2 2 ₂ 2 ₂

2 2

1 2 1 1 1 1 2 1 ₂ 1 2 2 ₁

4 2 1

.

w+ ≤C ∂ w+ ∂ w+ ≤C w+∂ w+ w+∂ w+ (15)

According to (7), we get

(

)

(

)

(

)

(

)

8 ₂ ₂ ₁ ₂ ₂ ₂ ₂ ₁ ₂ ₂

8 2 2 8 2 2

5

2 ₈ 2 ₈ 2 ₂ 2 ₂ .

p w w w w w w

w w w w

− + + + − −

− + − +

∇ ≤ ∂ + ∂ + ∂ + ∂

≤ + ∂ + ∂ (16)

(6)

(

)(

)

(

)

(

)

1 1

2 2

1 2 ₈ 2 ₈ 2 ₈ 2 ₂ 2 ₂ 1 2 1 ₂ 1 2 2 ₁

2 2 2

1 2 1 ₂ 2 ₂ 2 ₂

4

4 2 2

2 ₈ 2 ₈ 2 ₈ 1 ₂ 2 2 ₂

1 4

.

I C w w w w w w w w w

w w C w w

C w w w w w

+ − + − + + + + +

+ + − +

+ − + + +

≤ + ∂ + ∂ ∂ ∂

≤ ∂ + ∂ + ∂

+ + ∂

(17)

Therefore, recalling Theorem 1 and Sobolev embedding theorem, we get a global bound for w1 ₄

+ _.

(

)(

)

(

)

4 2 2

1 2 1 1

4

4 4 2 2

2 ₄ 2 ₄ 2 ₂ 2 ₂

4

4 2 2

2 ₈ 2 ₈ 2 ₈ 1 ₄ 2 2 ₂ d

d d

1

.

w w w x

t

C w w w w

C w w w w w

+ + +

− + − +

+ − + + +

+ ∂

≤ + + ∂ + ∂

+ + ∂

∫

Similarly, we can be established bound for 1 4

w− . To prove the 6

L -bound in

(8), we get from (3) that

(

)

(

)

2 2

6 6 2 2

1 ₆ 1 ₆ 1 2 1 1 2 1

2 2

4 4

1 1 1 1 1 1

1 d

5 5

6 d

5 d .

w w w w w w

t

p w w w w x

+ − + + − −

+ + − −

+ + ∂ + ∂

=

_∫

∂ + ∂

Note that

(

)

(

)

(

)

(

)

4 4

1 1 1 1 1 1

4 4

1 2 2 1 2 2

4 4

2 1 2 1 2

3 3

1 2 1 2 1 2 1 2

5 d

20 d .

p w w w w x

p w w w w w w x

+ + − − + + − −

+ + − −

+ + + − − −

∂ + ∂

= − ∂ + ∂

= ∂ +

+ ∂ + ∂

∫

Using Hölder’s inequality, (6) and Lemma 2, we obtain

(

)

(

) (

)

4 4

2 1 2 1 2

4 4

3 3 3 3

36

2 1 2 ₃₆ 1 2 ₃₆

3 3

19

2

2 ₃₆ 2 ₃₆ 2 ₂ 2 ₂

4 4 4 4 4 4

3 9 3 9 3 9 3 9 3 9 3 9

1 1 2 1 1 1 1 2 1 1

1 2 2 1 2 2

d

p w w w w x

p w w w w

C w w w w

w w w w w w

+ + − −

+ − + −

+ + + − − −

∂ +

 

 

≤ ∂ +

 

 

≤ + ∂ + ∂

 

 

× ∂ ∂ + ∂ ∂

 

 

∫

(

)

(

)

4 4

2 3 ₉ 3 ₉

2 ₃₆ 2 ₃₆ 2 ₂ 2 ₂ 1 1

2 2

4 4

8 4 8 4

3 9 3 9

9 9 9 9

1 ₄ 1 1 ₂ 1 ₄ 1 1 ₂ 2 1 2 1

2 2

.

C w w w w w w

w w w w w w

+ − + − + −

+ + − − + −

 

 

≤ + ∂ + ∂ +

 

 

 

 _ _

× ∂ + ∂ _ ∂ + ∂ _

 _ _

(7)

(

)

(

)

(

)

3 3

1 2 1 2 1 2 1 2

2 2

1 1 2 1 2 1 1 2 1 2

6 6 6 6 6

2 2

2 ₆ 2 ₆ 2 ₂ 2 ₂

2 2

1 ₆ 1 2 1 2 ₆ 1 ₆ 1 2 1 2 ₆

2 2

d

.

p w w w w w w x

p w w w w w w w w

C w w w w

w w w w w w w w

+ + + − − −

+ + + + − − − −

+ − + −

+ + + + − − − −

∂ + ∂

 

≤ _ ∂ + ∂ _

 

≤ + ∂ + ∂

 

×_ ∂ + ∂ _

 

∫

Therefore, by Young’s and Gronwall’s inequalities,

(

₆ ₆

)

₂ 2 ₂ 2

1 ₆ 1 ₆ ₀ 1 2 1 1 2 1

2 2

d .

t

w+ + w− + _ w+ ∂ w+ + w− ∂ w− _ τ≤C

 

∫

(18)

We now proved the inequality (9), taking the divergence of the first two equations in (3), we get

(

)

.

p w− w+

−∆ = ∇ ⋅ ⋅∇

Following the finiteness of Riesz transforms on p

L , we have

2 2 .

q q q

p ≤C w− w+

For 1< ≤q 3, according to Theorem 1 and (8),

2q

w+ and 2q

w− is

bounded, thus pq <C.

Recall that the operator _Λs

is defined through the Fourier transform [11], namely

s

_{( )}

s ˆ

_{( )}

_.

f ξ ξ f ξ

Λ =

Combining (6), Hardy-Littlewood-Sobolev inequality [12] and the boundedness of Riesz transforms in 2

L , we obtain

( )

(

)

( )

(

)

( )

(

)

( )

(

)

(

)

1

1 2 2 1 2 2 1

2 ₂

1

2 2 2 2 2 2 2 2

1 1

2 2 1 2 2 1 2 2 2 2 2 2

2 2

2 2 1 2 2 1 2 2 2 2 2 2

2 2

2 2 2 2

2 2

1 1

,

s s

s

s s

q q

s s

p w w w w

w w w w

w w w w w w w w

C w w w w w w w w

C w w w w

− − + + − − − + + −

− − − + + − − − − + + − − + + − − + + −

+ − + −

− −

Λ ≤ Λ −∆ ∂ ∂ + ∂

+ Λ −∆ ∂ ∂ + ∂

≤ Λ ∂ + ∂ + Λ ∂ + ∂

≤ ∂ + ∂ + ∂ + ∂

 

≤ ∂ + ∂ _ + _

 

(19)

with 1 1 1

2 2

s q

−

= + _{and C is a constant independent of s.}□

4. An Improved Global Lebesgue Bound

From the conclusions of Sections 2 and 3, we have the main theorem of this paper.

(

)

2

( )

2 0, 0

u b ∈H  be the corresponding solution

of (2). Let 2< < ∞r , then

(8)

DOI: 10.4236/ajcm.2019.92007 88 American Journal of Computational Mathematics Before proving the Theorem 3, we first describe the lemma that will be used. Lemma 3. [4] Let q∈

[

2,∞

)

and 1,1

2

s∈ _

 . Assume

( )

2 2

1

, ,

f g ∂ ∈g L  ,

( )

2q 1 2

h∈L −  and 2

( )

2

s_h _L

Λ ∈  . Then,

( )

2

1 1

1 2

2 2 2 2 1 2

d s ,

q

fgh x ≤C f g ρ ∂g −ρ hγ ₋ Λ h −γ

∫

 (21)

where ρ _and γ _{are given by}

(

)(

)

(

)(

)

(

)(

)

2 1 2 2 1 1

1

, ,

2 2 2 1 1 2 2 1 1 1

s q s q

ρ= + − − γ = − −

− − + − − + (22)

and 2

s

Λ _{denotes a fractional with respect to vertical dissipation and is defined by}

( )

2 2 e 2 ˆ d .

s

s ix

h x ξ ξ h ξ ξ

Λ =

_∫

(23)

Lemma 4. [8] Let s

( )

2

f ∈H  and B

(

0,R

)

denote the ball centered at

zero with radius R and by χB(0,R) the characteristic function on B

(

0,R

)

with

(

0,

)

R∈ ∞ and s∈

( )

0,1 . Write

( )

(

)

(

( )

)

1 1

0, 0,

with _B _R and _B _R ,

f = +f f f =− χ f f=− χ f (24)

where  and _−1 denote the Fourier transform and the inverse Fourier

transform. We have the following estimates for f and f .

1) For a pure constant C0 (independent of s)

( )

2

1 0

.

1 s

s H

C

f R f

s

−

∞ ≤ ₋  (25)

2) For any 2≤ < ∞q satisfying 1 s 2 0 q

− − < _,_{there is a constant} C₁

independent of s, q, R and f such that

( )

2

2 1

1 s .

s q

H q

f C qR f

− −

≤ 

 (26) Details can be seen in [8], we have omitted here.

Lemma 5. Let 1< < ∞q . Let q

( )

n

f ∈L  and let f be defined as in (24).

Then, there exists a constant C depending on q only such that

. q q

f ≤C f

Next we prove the Theorem 3.

Proof. According to Theorem 1, we have

(

)

(

)

2 2 2 2 2 2

2 ₂ 2 2 2 2 2 2

1 d

2 1 d 2 1 d ,

2 d

r r r

r

w r w w x r p w w x

r t

− −

+ ₊ ₋ _∂ + + ₌ ₋ _∂ + +

∫

(27)

with r>2. The right side of Equation (27) will be estimated using a different method. First, we fix R>0 and write

(

)

(

)

(

)

2 2 2 2 2

2 2 2 2

2 2 2 2 2 2

1 2

2 1 d

2 1 d 2 1 d

,

r

r r

r p w w x

r p w w x r p w w x

J J

− + +

− −

+ + + +

− ∂

= − ∂ + − ∂

= +

∫

(9)

DOI: 10.4236/ajcm.2019.92007 89 American Journal of Computational Mathematics where p and p as defined in (24). To estimate J1 and J2, let

(

)

5 1 5

1, 2 ,

2 2

3 1 1

1 .

2 2 2 1 1

s q

q

s s

− _{< <} _{< ≤}

+ < < +

− −

(28)

By Hölder’s and Young’s inequalities, we obtain

(

)

(

)

1 1

1 2 2 2 2

2 2

1 1

2

2 2 2 2

2 2

2 1

2 1 .

4

r r

J r p w w w

r

r p w w w

− −

+ + +

∞

− −

+ + +

∞

≤ − ∂

−

≤ − + ∂

Applying Lemma 4, we have

1 0

, 1

s s H

C

p R p

s

−

∞ ≤ ₋ (29)

where C0 is a constant independent of s. In the rest of the proof, we focus on

whether a constant is bounded uniformly as s→1−. Using the interpolation inequality, we have

( )

2 22 4

2 2

1 1

2 d 2 ₂ 2 ₂ .

r r r

r r

r

w x w w

− −

+ _≤ + ₋ + ₋

∫

(30)

In summary, we obtain

( )

1 2 02

(

)

2 1( ) 2 21 2214

1 2 2 2 2 ₂ 2 ₂

2

2 1

2 1 ,

4 1 s

r r

r _s

r r

H r

C r

J w w r R p w w

s

−

− ₋

+ + + − + −

−

≤ ∂ + −

− (31)

where C0 is independent of s. Now we estimate J2, apply Lemma 3 to obtain

(

)

1 ₍ ₎ 1 1

( )

11

2 2 2 2 2 1 1 2 2 2 2

2 2 2

2 1 r _q s r r ,

J C r w w p p w w

ρ ρ

γ

γ − −

− − −

+ + + +

−

≤ − ∂  Λ  ∂

where s and q satisfy (28), γ _and ρ _{are given explicitly in terms of s and q}

(

)(

)

(

)(

)

(

)(

)

2 1 1 1 2 1 2

, ,

2 1 1 1 2 2 2 1 1 1

s q s q

γ = − − ρ= + − −

− − + _ − − + _ (32)

and C is bounded uniformly as s→1−. According to (30), we get

(

2 ₂

)

1

1 1

2 2 ₂ 2 ₂

2

. r r r

r r

r

w w w

ρ ρ

ρ −

−

+ _≤ + − + − ₍₃₃₎

By Hölder’s inequality,

( )

(

)

(

) ( )

( )

)

( )

(

)

(

) (

)

( )

(

)

(

( )

(

)

( )( ) ( )

1 1 1

1 ₁ 2 2 2 ₂

2 2 2 2 2

2

1

2 2

1

1 1

2 2 2 2 2

2 1 1

2 1 2

1 2 1

1

2 2 ₂ 2 2 2

1 d

1 d .

r r

r

r r

r

r r

r

w r w w x

r w w w x

ρ ρ _ρ

ρ ρ

ρ

− −

− − −

+ + +

− −

−

− + ₋ + ₋ +

− −

− ₋

− + − + + −

∂ = − ∂

 

= − _ ∂ ∂ _

 

= − ∂ ∂

∫

(10)

(

)(

)

(

)

( )

(

) ( )

)

( )( ) ( )

(

) ( )

(

)(

)

( )( ) ( )

(

)

( ) ( ) ( )( ) 2 2 2 1 1

1 1 1

2 2 2 ₂ 2 ₂ 2 ₂

2 1 1

2 2 2

1 ₂ ₂ ₁

2 2 2

2 1 2

2 2 1 1

2 2 2

2 2 2 2 ₂

2 2 2 1

2 1 2 1 1

2 2 2 2 2 2 1 2

2 1 1

d

2 1

d 2 1 1

4

,

r r

r r r

r r r _r s q r r

r r r r

s q r

J C r r w w w

p p w w x

r

w w x C r r w

w w p p

ρ ρ ρ ρ ρ γ γ ρ ρ σ σ ρ γ ρ γ σ

σ σ σ

− − − + ₋ + ₋ + ₋ − − + − − ₊ ₊ ₋ − − − − + + + − ₋ − − − + + − ≤ − − ∂

× Λ ∂

−

≤ ∂ + − −

× ∂ Λ

∫

 

(34)

where C is again bounded uniformly as s→1−, and we make

(

r 1

) (

1

)(

r 2

)

1 r 2 .

σ = − − −ρ − = +ρ − ρ ₍₃₅₎

For further estimation, we spilt p 2₍q−1₎ into two parts and bound one of them by Lemma 4. Moreover, we get any 0≤ ≤β 1,

( ) ( ) ( ) ( )

( )

1 1

1 1 1

1

2 1 2 1 2 1 2 1

1 1

2 1 .

s q

s

q q q q H

s q

s

q H

p p p C p R p

C p R p

β

β β β β

β β β _{− −}    − −  − − − − − _{− −}    −  − − = ≤ ≤     (36)

Owing to the condition of s and q in (28), this boundary allows us to generate

1 1 1 s q R β _{− −}   ₋

  _with ₁ 1 ₀

1 s q β   − − ≤  ₋ 

  . Inserting (36) in (34) yields

(

) ( )

(

)(

)

( )( ) ( ) ( )

(

)

( ) ( ) ( ) ( ) ( )( ) 2

2 2 2

2 2 2 2

2 1

1 ₂

2 1 1 1

1

2 ₂ 2 1

2 2 ₂ ₁ _{2 1} ₁

2 1 ₁

2 2 2 2 2 2 1

2 1

d 4

2 1 1

, s r r r s q r

r r _r _r

q H

r

J w w x

C r r R w

w w p p

γ _ρ

ρ β

σ _σ σ

γ

ρ γ γ

ρ _β _β

σ _σ _σ σ σ − + + −   − − − − ₋ ₊   −   − ₋ ₋ ₋ − _{− } _ ₊   + +   − − ≤ ∂ + − − × ∂

∫

where C is bounded uniformly as s→1−. We choose β so that the sum of the powers of 2w2 ₂

+

∂ _{and of} p _Hs is equal to 2, namely

(

)

(

)

(

)(

)

2 1 2 1 2 1 1

2.

r r

ρ γ γ

β

σ σ σ

− − − −

+ + =

Recalling (32) and (35), we have

(

)(

)

(

)(

)

2 1 2 3 1

.

2 2 2 1

s q

q s

β = − − −

− − (37)

The condition in (28) ensures that 0< ≤β 1, then

( ) ( ) ( )( )

(

)

2 1 2 1 1 2 1

2 2

2 2 ₂ s 2 2 ₂ s .

r r

H H

w p C w p

γ γ ρ _β σ σ σ − − − − + + + ∂ ≤ ∂ +

For β _{given by (37), we have}

(

) ( )

(

)(

)

( )( ) ( )

( )

( ) ( )

(

)

(

2

)

2 2 2

2 2 2 2

2 1

1 ₂

2 1 1 1

1

2 2

2 1

1 2 2

2 2 2

2 1 2 2

2 1

d 4

2 1 1

. s r r r s q r r r

q H r

r

J w w x

C r r R w

p w p w

(11)

DOI: 10.4236/ajcm.2019.92007 91 American Journal of Computational Mathematics Combining (27), (31) and (38) we have

(

)

( )

(

)(

)

( )( ) ( ) ( ) ( ) ( )

(

)

(

)

2 2

2 2 2 2

2 ₂ 2 2 2

2 2 2 4

2 2 1

0 1 1

2 ₂ 2 ₂

2 1

1 ₂

2 1 1 1

1

2 ₂

2 2

2 1

1 2 2

2 2 2

2 1 2 2

1 d 2 1

d

2 d 4

2 1

1

2 1 1

, s s r r r r r

s _r _r

H r r r s q r r r

q H r

r

w w w x

r t

C

r R p w w

s

C r r R w

p w p w

γ _ρ ρ β σ _σ σ ρ γ β σ σ − + + + − − + ₋ + ₋ −   − − − − ₋  ₊   − − − ₊ ₊ − − + ∂ ≤ − − + − − × ∂ +

∫

(39)

with a constant C0 is independent in s and C is bounded uniformly as s 1

−

→ _.

Let

( )

₍

₎

2 1( )( 1) 1 1 2( 1)

2 1 1

1 ,

r

r _s

s q

R r R

γ ρ _β σ σ −   − − __{− −} _ − ₌ ₋  − that is, ( )

₍

₎

( )( )( ) ( ) ( )

2 1 1 1 1

1 1 1

2 1 ₁

1

s r

s s r

s _q R r ρ σ βγ − − −   − +  − +  −

− ₌ ₋ _ ₋_ ₍₄₀₎

Using (32), (35) and (37) to simplify this index and get

(

)(

)

(

)

(

)

(

)(

)

(

) (

1

)(

)

2 1 1 1 2 1 1

.

1 2 1 1 1

1 1 1

1

s r s q

q r s q

s s r

q

ρ

σ βγ −

− − − − − =   − + − − − − + _ − + _ − −   Let

(

)(

)

(

) (

1

)(

)

2 1 1

,

2 1 1 1

s q

q r s q

θ = − ₋ −

− + − − − (41)

and therefore 2 1( )

(

)

1 s

R − = r− θ. Obviously, θ→0 as s→1, and

(

)

1 1 1 2 2

2 .

1 2 1 2

q q

s q r q

θ

θ θ

−   −

= _ − _ ≤

− −  −  − (42)

Furthermore,

(

2

)

2 ₂ ₂

2 4

2 2, 2 2.

1 r r r r r r r ρ σ − − _≤ ₋ _≤ ₋

− (43)

For generality, we assume w2 ₂_r 1

+ _≥

. Following (39) and get

( )(

)(

)

2 2 ₂ d

2 1 1 ,

d r

C

w A t r r

t

θ θ

+ _≤ ₋ ₋ ₍₄₄₎

where C is bounded uniformly as θ→0+, and

( )

2 2 2 (1₍ ) ( )₎2 1

(

2 2

)

1

2 2 2 2 2 1 2 2 2 .

s s

r r

H q H

A t p w w p w p

γ ρ _β σ σ − − + − + + − = + ∂ +

Since (44) holds for any θ>0, we set

(

1

)

,

log r 1

(12)

DOI: 10.4236/ajcm.2019.92007 92 American Journal of Computational Mathematics we get

( )(

) (

)

2 2 ₂ d

2 1 log 1 .

d r

C

w A t r r

t θ

+ _≤ ₋ ₋ ₍₄₅₎

Choose the right θ, and according to Theorem 3, A t

( )

is integrable at any

time interval. This completes the proof of Theorem 3. □

5. Conditional Global Regularity

This section estimates the global boundedness of the vertical component u2

and b2 of

( )

, 2

H

u b under the L L2t y

∞_{norm. We have the following theorem.}

Theorem 4. Assume

(

)

2

( )

2

0, 0

u b ∈H  and

( )

u b, be the corresponding solution of (2). If

(

)( )

2

2 2

0 , d

T

u b τ _∞ τ < ∞

∫

for some T >0, then

( )

, 2

H

u b is finite on

[ ]

0,T .

We divide the proof of the theorem into two parts.

5.1. H1_{in Terms of}

(

)

t y

L L

u b2, 2 2∞

In this section, we estimate that the solution has a 1

H -bound, and we have the

following proposition.

Proposition 5. Assume

(

)

2

( )

2

0, 0

( )

u b, be the corresponding solution of (2). Then, for any T >0 and t≤T,

( )( )

20

(

2( )2 2( )2

)

1

d 1

, e ,

t

C u b

H

u b t ≤C ∫ τ ∞+ τ ∞ τ (46)

where C1 depends on T and the initial data only and C2 is a pure constant.

Proof. Taking the inner product of the first equation of (3) with ∆w+ and

integrating by parts, we find

2 2

2 1 2 3 4 5 6

2 2

1 d

,

2 dt w w I I I I I I

+ +

∇ + ∂ ∇ = + + + + +

where

1 1 1 1 2 1 2d , 2 1 2 2 1 1 1d ,

I = ∂

∫

w−∂w+∂w+ x I = ∂

∫

w−∂ w+∂w+ x

3 1 2 2 2 1 2d , 4 2 1 1 1 2 1d ,

I = ∂

∫

w−∂ w+∂ w+ x I = ∂

∫

w−∂ w+∂ w+ x

5 2 1 1 2 2 2d , 6 2 2 2 1 2 1d .

I = ∂

∫

w−∂ w+∂ w+ x I = ∂

∫

w−∂ w+∂ w+ x

Using the anisotropic Sobolev inequalities [5] and ∇ ⋅w+ = ∇ ⋅w−=0, we can be bounded as follows,

1 2 2 1 2 1 2 2 1 2 12 2

2 1 2 ₂ 12 2 ₂

2 2 2

2 2 2 1 2

2 2

d

2 d

1

, 8

I w w w x

w w w x

C w w w

w C w w

− + + − + + − + +

∞

+ − + ∞

= − ∂ ∂ ∂

= ∂ ∂

≤ ∂ ∂

≤ ∇∂ + ∂

(13)

2 1 2 2 1 1 1

1 1 1 1

2 2 2 2

1 2 2 1 12 1 1 1 12 1

2 2 2 2 2

1 1

2 2

2 2 1 12 1 2 2

2 2 2 2

2 ₂ 2 ₂ 2 1 ₂

2 2 2

2 1 2 2

2 2 2

d

1

, 8

I w w w x

C w w w w w

C w w w w

C w w w

w C w w

− + +

− + + + +

− + + +

− + +

+ − +

= ∂ ∂ ∂

≤ ∂ ∂ ∂ ∂ ∂

= ∇ ∂ ∂ ∂

≤ ∇ ∂ ∇∂

≤ ∇∂ + ∇ ∂

∫

(

)

3 1 2 2 2 1 2

12 2 2 1 2 1 2 2 12 2

2 12 2 ₂ 2 2 ₂ 1 2 ₂ 12 2 ₂

2 2 2 2 2 2

2 2 ₂ 2 2 2 ₂ 2 2 ₂ 2 1 2 ₂

d

d d

1 1

,

8 8

I w w w x

w w w x w w w x

C w w w w w

w C w w w C w w

− + +

− + + − + +

+ − + − +

∞

− + + + + −

∞ ∞

= ∂ ∂ ∂

≤ − ∂ ∂ + − ∂ ∂

≤ ∂ ∂ + ∂ ∂

≤ ∇∂ + ∂ + ∇∂ + ∂

∫

4 2 1 1 1 2 1

1 1 1 1

2 2 2 2

2 1 ₂ 1 1 ₂ 12 1 ₂ 2 1 ₂ 12 1 ₂

2 1 ₂ 1 ₂ 2 1 ₂

2 2 2

2 1 2 1 1

2 2 2

d

1

, 8

I w w w x

C w w w w w

C w w w

w C w w

− + +

− + + + +

− + +

+ − +

= ∂ ∂ ∂

≤ ∂ ∂ ∂ ∂ ∂

≤ ∂ ∇ ∇∂

≤ ∇∂ + ∂ ∇

∫

5 2 1 1 2 2 2

1 1 1 1

2 2 2 2

2 1 1 2 12 2 2 2 12 2

2 2 2 2 2

2 1 ₂ 2 ₂ 2 2 ₂

2 2 2

2 2 ₂ 2 1 ₂ 2 ₂

d

1

, 8

I w w w x

C w w w w w

C w w w

w C w w

− + +

− + + + +

− + +

+ − +

= ∂ ∂ ∂

≤ ∂ ∂ ∂ ∂ ∂

≤ ∂ ∇ ∇∂

≤ ∇∂ + ∂ ∇

∫

6 2 2 2 1 2 1

1 1 1 1

2 2 2 2

2 2 ₂ 2 1 ₂ 12 1 ₂ 2 1 ₂ 12 1 ₂ 2 2 ₂ 2 1 ₂ 2 1 ₂

2 2 2

2 1 ₂ 2 2 ₂ 1 ₂ d

.

1

. 8

I w w w x

C w w w w w

C w w w

w C w w

− + +

− + + + + − + +

+ − +

= ∂ ∂ ∂

≤ ∂ ∂ ∂ ∂ ∂

≤ ∂ ∂ ∇∂

≤ ∇∂ + ∂ ∇

∫

Similarly, we can estimate ∇w−. Combining them yields

(

) (

)

(

)(

)

2 2 2 2

2 2

2 2 2 2

2 2 2 2 2 2

2 ₂ 2 ₂ 2 2 ₂ ₂

d d

.

w w w w

t

C w w w w w w

+ − + −

+ − + − + −

∞ ∞

∇ + ∇ + ∇∂ + ∇∂

≤ ∂ + ∂ + + ∇ + ∇ (47)

According to Gronwall’s inequality, get

(

∇ ∇u, b

)

has a L2-bounded.

Combining with the Lemma 1 to got (46). □

5.2. Proof of Theorem 4

(14)

DOI: 10.4236/ajcm.2019.92007 94 American Journal of Computational Mathematics completion of the Theorem 4.

Proof. Taking the inner product of the first equation in (3) with 2 w+

∆ _and

integrating by parts, we find

(

)

2 2

2

2 2

1 d

d .

2 dt w w w w w x

+ + − + +

∆ + ∂ ∆ = − ∆

_∫

⋅∇ ⋅ ∆ ₍₄₈₎

We decompose the nonlinear term into different parts and estimate it using anisotropic dissipation. We write

(

w w

)

w dx K1 K2 K3,

− + +

∆ ⋅∇ ⋅ = + +

∫

with

(

)

(

)

1 d , 2 2 1 1 d ,

K =

∫

∆w−⋅∇w+ ⋅ ∆w+ x K =

∫

∂ w−⋅∇∂w+ ⋅ ∆w+ x

(

)

3 2 2 2 d .

K =

∫

∂ w−⋅∇∂ w+ ⋅ ∆w+ x

We further divide K1 into four parts, K1=K11+K12+K13+K14, where

(

)

(

)

11 1 1 1 1d , 12 1 1 2 2d ,

K =

∫

∆ ∂w− w+ ∆w+ x K =

∫

∆ ∂w− w+ ∆w+ x

(

)

(

)

13 2 2 1 1d , 14 2 2 2 2d ,

K =

∫

∆ ∂w− w+ ∆w+ x K =

∫

∆ ∂w− w+ ∆w+ x

Applying Hölder’s inequality and ∇ ⋅w+=0, after integration by parts we get

(

)

(

)

(

)

11 1 2 2 1

2 1 2 1 1 2 2 1

2

2 2 1 1 2 2 1 1

2 2 2 2

2 2 2 2 2

2 1 ₂ 2 1 ₂ 2 1 ₂ 1 ₂

d

d d

1

. 48

K w w w x

w w w x w w w x

C w w w C w w w

w w C w w w

− + +

− + + − + +

+ − + + + −

∞ ∞

− + + + −

∞

= − ∆ ∂ ∆

≤ ∆∂ ∆ + ∆ ∆∂

≤ ∆∂ ∆ + ∆∂ ∆

≤ ∆∂ + ∆∂ + ∆ + ∆

∫

Similarly, we obtain

(

2 2

)

2

(

2 2

)

14 2 2 ₂ 2 2 ₂ 2 2 ₂ 2 ₂

1

. 48

K ≤ ∆∂ w− + ∆∂ w+ +C w+ _∞ ∆w+ + ∆w−

To bound K12 and K13 , we use anisotropic Sobolev inequality and

Proposition 5, we obtain

1 1 1 1

2 2 2 2

12 2 ₂ 1 ₂ 1 1 ₂ 1 2 ₂ 12 2 ₂

1 1 1 1

2 2 2 2

2 ₂ 1 ₂ 2 2 ₂ 2 ₂ 2 2 ₂

2 2 2 2

2 2 ₂ 2 2 ₂ 1 ₂ 2 ₂ 2 ₂

1

, 48

K C w w w w w

C w w w w w

w C w w C w w

+ − − + +

− + − + +

≤ ∆ ∆ ∆∂ ∂ ∂

≤ ∆ ∆ ∆∂ ∇ ∇∂

≤ ∆∂ + ∇∂ ∆ + ∇ ∆

(

)

(

)

1 1 1 1

2 2 2 2

13 2 1 ₂ 2 ₂ 2 2 ₂ 1 ₂ 1 1 ₂

1 1 1 1

2 2 2 2

2 1 ₂ 2 ₂ 2 2 ₂ 1 ₂ 2 2 ₂

2 2 2 2 2

2 2 2 2 2 1 2 1

2 2 2 2 2

1

. 48

K C w w w w w

C w w w w w

w w C w w w

+ − − + +

− + + − +

≤ ∂ ∆ ∆∂ ∆ ∆∂

≤ ∆∂ + ∆∂ + ∂ ∆ + ∆

(15)

(

) (

)(

)

2 2 2 2

1 2 ₂ 2 ₂ 2 2 ₂ 2

2 2 2

2 ₂ 2 ₂ ₂ ₂

1 12

.

K w w C w w

w w w w

− + + +

∞

+ + + −

≤ ∆∂ + ∆∂ + ∇∂ +

+ ∇ + ∂ ∆ + ∆

2

K and K3 can be estimated in a similar way and here we will omit the

details. Combining all of these estimates and applying Gronwall’s inequality, we have

(

) (

)

2 2

2

2 2

2 2 2 2

2 2 1

2 2 2 2

1 d 2 d

1

, 4

w w

t

w w B w w

+ +

− + + −

∆ + ∆∂

≤ ∆∂ + ∆∂ + ∆ + ∆ (49)

where

2 2 2

1 2 2 ₂ 2 2 ₂ 2 ₂.

B w+ w+ w+ w+

∞

= ∇∂ + + ∇ + ∂

Similarly,

(

) (

)

2 2

2

2 2

2 2 2 2

2 2 2

2 2 2 2

1 d 2 d 1

, 4

w w

t

w w B w w

− −

− + − +

∆ + ∆∂

≤ ∆∂ + ∆∂ + ∆ − ∆ (50)

and

2 2 2

2 2 2 ₂ 2 2 ₂ 2 ₂.

B w− w− w− w−

∞

= ∇∂ + + ∇ + ∂

combines with (50) and (49), we get

(

) (

)

(

)

(

)

2 2 2 2

2 2

2 2 2 2

2 2

1 2 ₂ ₂

d d

.

w w w w

t

B B w w

+ − + −

− +

∆ + ∆ + ∂ ∆ + ∂ ∆

≤ + ∆ − ∆

Applying Gronwall’s inequality and (4), (8), (47), we can prove that the solution

( )

u b, in (2) has a global H2-bound. This completes the proof of

Theorem 4. □

6. Conclusion

According to Wu [4], in this paper, we prove that the solution of the system (2) has regularity in the vertical direction. In order to get this result, we need to make a corresponding estimate of the pressure to prove that it’s bounded. Especially in the case of

(

)

2

2 2

0 , d

T

u b _∞ t< ∞

∫

, the solution has regularity in

[ ]

0,T .

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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